The answer is no, of course, since Cos(0) = 1, i*Sin(0) = 0. Further, there are clearly no solutions to this equation for real x, since the cosine will be real, and the sine imaginary. But are there solutions for complex x?

Cos(a+bi) = i*Sin(a+bi)

let's start with the cosine term, and break it up using the angle addition law

Cos(a + bi) = Cos(a)Cos(bi) - Sin(a)Sin(bi)

We now need the sines and cosines of imaginary angles. We can get them using Euler's equation.

e

^{ix}= Cos(x) + i*Sin(x)

e

^{-ix}= Cos(-x) + i*Sin(-x) = Cos(x) - i*Sin(x) using the even and odd properties of sine and cosine

now add these

e

^{ix}+ e

^{-ix}= 2Cos(x)

let x be an imaginary number i*b, b is real

e

^{-b}+ e

^{b}= 2Cos(i*b) = 2Cosh(b)

where the last bit about the hyperbolic cosine is just from the definition Cosh(x) = [e

^{x}+ e

^{-x}]/2

do the same to find the sine of an imaginary angle. you'll obtain

Sin(i*b) = i*Sinh(b)

go back and plug this into the expansion of Cos(a+bi)

Cos(a+bi) = Cos(a)Cosh(b) - i*Sin(a)Sinh(b)

Sin(a+bi) = Sin(a)Cosh(b) + i*Sinh(b)Cos(a)

our original equation was iSin(x) = Cos(x), with "x" complex ("a" and "b" above are real). Taking the two equations above and equating their real and imaginary parts

Cos(a)Cosh(b) = -Sinh(b)Cos(a)

-Sin(a)Sinh(b) = Sin(a)Cosh(b)

Both of these equations yield the same condition:

Cosh(b) = -Sinh(b)

Tanh(b) = -1

But Tanh(b) approaches this value asymptotically in the limit b ==> -Infinity

So there are no solutions to the equation.

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